I'm trying to diagonalise the matrix $A = \begin{pmatrix}3&-1&1\\0&2&2\\1&-1&3 \end{pmatrix}$. I know that the answer is $D = \begin{pmatrix}2&0&0\\0&2&0\\0&0&4 \end{pmatrix}$ but I'm trying to understand the full process of getting there.
I found a basis for A consisting of eigenvectors which is $\left\{\begin{pmatrix} 1\\ 1\\ 0 \end{pmatrix}, \begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}, \begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}\right\}$.
So using $P^{-1}AP=D$, we have $P = \begin{pmatrix}1&-1&1\\1&0&0\\0&1&1 \end{pmatrix}$ and $P^{-1} = \begin{pmatrix}0&1&0\\-1/2&1/2&1/2\\1/2&-1/2&1/2 \end{pmatrix}$.
Then $D=\begin{pmatrix}0&1&0\\-1/2&1/2&1/2\\1/2&-1/2&1/2 \end{pmatrix}\begin{pmatrix}3&-1&1\\0&2&2\\1&-1&3 \end{pmatrix}\begin{pmatrix}1&-1&1\\1&0&0\\0&1&1 \end{pmatrix}$
$\implies D=\begin{pmatrix}4&0&0\\1&2&0\\1&0&2 \end{pmatrix}$ which clearly isn't what I want. Where did I mess up?
Note that $v_1:=\begin{pmatrix} -1\\ 0\\ 1 \end{pmatrix}$ and $v_2:=\begin{pmatrix} 1\\ 0\\ 1 \end{pmatrix}$ are not eigenvectors of $A$ because $$Av_1=-2v_1+\begin{pmatrix} 0\\ 2\\ 0\end{pmatrix}\qquad\text{ and }\qquad Av_2=4v_1+\begin{pmatrix} 0\\ 2\\ 0\end{pmatrix}.$$ In fact the matrix is not diagonalizable over $\Bbb{R}$ because its characteristic polynomial $$P_A(X)=X^3-8X^2+22X-20=(X-2)(X^2-6X+10),$$ has nonreal roots, because $X^2-6X+10$ is irreducible over $\Bbb{R}$.